The equilibrium constant is 8.8 x 102 at a particular temperature for the reacti

Question

The equilibrium constant is 8.8 x 102 at a particular temperature for the reaction: H2(g) + I2(g) 2HI(g) Use this information to decide what will happen, given the following sets of initial conditions: Will the reaction shift to the left, to the right or will the system be at equilibrium?

1. A 3.0 L flask contains 1.17 mol H2, 0.075 mol I2 and 8.39 mol HI.

2. A 3.0 L flask contains 0.0601 mol H2, 0.009 mol I2 and 0.69 mol HI.

3 . PH2 = 0.45 torr, PI2 = 0.062 torr, PHI = 6.11 torr

4. PH2 = 0.5627 torr, PI2 = 0.067 torr, PHI = 5.76 torr

Explanation / Answer

there is equal number of particles on reactant and product.
So we can work with number of moles rather than concentration as volume willcancel each other in both numerator and denominator
Kc = Kp = 8.8*10^2 = 880

1)
Q = (8.39)^2 / (1.17 * 0.075)
= 802.2
since Q < K, reaction will shift to right

2)
Q = (0.69)2 / (0.0601 * 0.009)
= 880
since Q is equal to K, system will be at equilibrium

3)
Q = (6.11)^2 / (0.45 * 0.062)
= 1338
since Q > K, reaction will shift to left

4)
Q = (5.76)^2 / (0.5627 * 0.067)
= 880
since Q is equal to K, system will be at equilibrium

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.